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Core Connections Algebra 1, 2013

CC

Core Connections Algebra 1, 2013 View details

2. Section 4.2

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Exercise 81 Page 176

a We will use the Elimination Method to solve this system of equations. It is usually the best choice when one of the variables has equal or opposite coefficients as they are in the given equation.

{- 2 x + 3 y = 1 2 x + 6 y = 2 (I) (II)

$(I):$ Add $(II)$

{- 2 x + 3 y + (2 x + 6 y) = 1 + 2 2 x + 6 y = 2

$(I):$ Remove parentheses

{- 2 x + 3 y + 2 x + 6 y = 1 + 2 2 x + 6 y = 2

$(I):$ Add terms

{9 y = 3 2 x + 6 y = 2

$(I):$ $LHS / 9 = RHS / 9$

{y = \frac{3}{9} 2 x + 6 y = 2

$(I):$ $\frac{a}{b} = \frac{a / 3}{b / 3}$

{y = \frac{1}{3} 2 x + 6 y = 2

Having found

y,

we can substitute this into the second equation to find

x .

{y = \frac{1}{3} 2 x + 6 y = 2 (I) (II)

$(II):$ $y = \frac{1}{3}$

{y = \frac{1}{3} 2 x + 6 (\frac{1}{3}) = 2

MoveLeftFacToNumOne

$(II):$ $a \cdot \frac{1}{b} = \frac{a}{b}$

{y = \frac{1}{3} 2 x + \frac{6}{3} = 2

$(II):$ Calculate quotient

{y = \frac{1}{3} 2 x + 2 = 2

$(II):$ $LHS - 2 = RHS - 2$

{y = \frac{1}{3} 2 x = 0

$(II):$ $LHS / 2 = RHS / 2$

{y = \frac{1}{3} x = 0

We can check our solution by substituting

x = 0

and

y = \frac{1}{3}

into the original system of equations. If the left-hand side and right-hand side are equal in both equations, the solution is correct.

{- 2 x + 3 y = 1 2 x + 6 y = 2 (I) (II)

$(I), (II):$ $x = 0$ , $y = \frac{1}{3}$

{- 2 (0) + 3 (\frac{1}{3}) = ? 1 2 (0) + 6 (\frac{1}{3}) = ? 2

$(I), (II):$ Multiply

{0 + 1 = ? 1 0 + 2 = ? 2

$(I), (II):$ Add terms

{1 = 1 2 = 2

Both equations are true, so our solution is correct!

b We will use the Substitution Method to solve this system of equations. It is usually the best choice when one of the variables is already isolated or has a coefficient of

1

or

- 1 .

In the second equation,

x

is already solved for so we can substitute it in the first equation to find

y .

{y = \frac{1}{3} x + 4 x = - 3 y (I) (II)

$(I):$ $x = - 3 y$

{y = \frac{1}{3} (- 3 y) + 4 x = - 3 y

FracMultDenomToNumber

$(I):$ $\frac{a}{3} \cdot 3 = a$

{y = - y + 4 x = - 3 y

$(I):$ $LHS + y = RHS + y$

{2 y = 4 x = - 3 y

$(I):$ $LHS / 2 = RHS / 2$

{y = 2 x = - 3 y

Having found

y,

we can substitute this into the second equation to find

x .

{y = 2 x = - 3 y

$(II):$ $y = 2$

{y = 2 x = - 3 (2)

$(II):$ Multiply

{y = 2 x = - 6

We can check our solution by substituting

x = - 6

and

y = 2

into the original system of equations. If the left-hand side and right-hand side are equal in both equations, the solution is correct.

{y = \frac{1}{3} x + 4 x = - 3 y (I) (II)

$(I), (II):$ $x = - 6$ , $y = 2$

{2 = ? \frac{1}{3} (- 6) + 4 - 6 = ? - 3 (2)

$(I), (II):$ Multiply

{2 = ? - 2 + 4 - 6 = - 6

$(I):$ Add terms

{2 = 2 - 6 = - 6

Both equations are true, so our solution is correct!

c Since the second equation is in slope-intercept form, let's rewrite the first equation in this form as well.

{3 x - y = 7 y = 3 x - 2 (I) (II)

$(I):$ $LHS - 3 x = RHS - 3 x$

{- y = - 3 x + 7 y = 3 x - 2

$(I):$ $LHS \cdot (- 1) = RHS \cdot (- 1)$

{y = 3 x - 7 y = 3 x - 2

Having rewritten the first equation, we see that they have the same slope but different

y -

intercepts. This makes them parallel lines, which means that they will never intersect. Thus, this system of equations does not have a solution.

d If we multiply the first equation by

3,

the coefficient of the

x -

variables will be the same. This means we can use the Elimination Method to solve the system.

{x + 2 y = 1 3 x + 5 y = 8 (I) (II)

$(I):$ $LHS \cdot 3 = RHS \cdot 3$

{3 x + 6 y = 3 3 x + 5 y = 8

$(I):$ Subtract $(II)$

{3 x + 6 y - (3 x + 5 y) = 3 - 8 3 x + 5 y = 8

$(I):$ Distribute $(- 1)$

{3 x + 6 y - 3 x - 5 y = 3 - 8 3 x + 5 y = 8

$(I):$ Subtract terms

{y = - 5 3 x + 5 y = 8

Having found

y,

we can substitute it in the second equation to find

x .

{y = - 5 3 x + 5 y = 8

$(II):$ $y = - 5$

{y = - 5 3 x + 5 (- 5) = 8

$(II):$ $a (- b) = - a \cdot b$

{y = - 5 3 x - 25 = 8

$(II):$ $LHS + 25 = RHS + 25$

{y = - 5 3 x = 33

$(II):$ $LHS / 3 = RHS / 3$

{y = - 5 x = 11

We can check our solution by substituting

x = 11

and

y = - 5

into the original system of equations. If the left-hand side and right-hand side are equal in both equations, the solution is correct.

{x + 2 y = 1 3 x + 5 y = 8 (I) (II)

$(I), (II):$ $x = 11$ , $y = - 5$

{11 + 2 (- 5) = ? 1 3 (11) + 5 (- 5) = ? 8

$(I), (II):$ Multiply

{11 - 10 = ? 1 33 - 25 = ? 8

$(I), (II):$ Subtract term

{1 = 1 8 = 8

Both equations are true, so our solution is correct!

Subchapter links

4.2.1 Solving Systems of Equations Using Substitution p.156-157

4.2.2 Making Connections: Systems, Solutions, and Graphs p.162-163

4.2.3 Solving Systems Using Elimination p.168-169

4.2.4 More Elimination p.173

4.2.5 Choosing a Strategy for Solving Systems p.176-177